Question

giải pt:

\(\left\{{}\begin{matrix}x^3+3y=y^3+3x\\x^2+2y^2=1\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+xy+y^2\right)+3\left(x-y\right)=0\\x^2+2y^2=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+xy+y^2+3\right)=0\\x^2+2y^2=1\end{matrix}\right.\)

Ta có \(x^2+xy+y^2+3=\left(x^2+2\cdot x\cdot\frac{1}{2}y+\frac{1}{4}y^2\right)+\frac{3}{4}y^2+3>0\)

\(\Rightarrow x-y=0\Rightarrow x=y\)

Thay vào phương trình

\(x^2+2y^2=1\Rightarrow x^2+2x^2=1\Rightarrow3x^2=1\Rightarrow x^2=\frac{1}{3}\)

\(\Rightarrow\left[{}\begin{matrix}x=y=\sqrt{\frac{1}{3}}\\x=y=-\sqrt{\frac{1}{3}}\end{matrix}\right.\)